Let $$S$$, be the set of all real roots of the equation, $$3^x(3^x - 1) + 2 = |3^x - 1| + |3^x - 2|$$, then S

We begin by observing that the given equation involves the expression $$3^x$$ inside absolute value signs. To simplify the subsequent algebra, let us set $$y = 3^x$$. Because an exponential is always positive, we automatically have $$y > 0$$.

With this substitution, the original equation

$$3^x(3^x - 1) + 2 = |3^x - 1| + |3^x - 2|$$

becomes

$$y(y - 1) + 2 \;=\; |y - 1| + |y - 2|.$$

Now the entire problem is to solve this equation for positive real $$y$$ and then back-substitute to obtain $$x$$. Because of the absolute values, we split the analysis into intervals decided by the critical points $$1$$ and $$2$$.

Case I: $$0 < y < 1$$ (both $$y-1$$ and $$y-2$$ are negative)

For this range

$$|y - 1| = 1 - y,\qquad |y - 2| = 2 - y.$$

Substituting into the equation gives

$$y(y - 1) + 2 \;=\; (1 - y) + (2 - y).$$

The right side simplifies to $$3 - 2y$$, while the left side expands to $$y^2 - y + 2$$. Hence

$$y^2 - y + 2 = 3 - 2y.$$ $$y^2 - y + 2 - 3 + 2y = 0.$$ $$y^2 + y - 1 = 0.$$

This is a quadratic in standard form. Using the quadratic formula $$y = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\; b = 1,\; c = -1$$, we obtain

$$y = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}.$$

Only the plus sign yields a positive value:

$$y = \frac{\sqrt5 - 1}{2}\approx 0.618.$$

Because $$0 < 0.618 < 1$$, this root indeed lies inside the assumed interval, so it is valid.

Case II: $$1 \le y < 2$$ (now $$y-1 \ge 0$$ but $$y-2<0$$)

Here

$$|y - 1| = y - 1,\qquad |y - 2| = 2 - y.$$

Substituting gives

$$y(y - 1) + 2 = (y - 1) + (2 - y) = 1.$$ $$y^2 - y + 2 = 1.$$ $$y^2 - y + 1 = 0.$$

The discriminant is $$(-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0,$$ so there is no real root in this interval.

Case III: $$y \ge 2$$ (both $$y-1$$ and $$y-2$$ are non-negative)

Now

$$|y - 1| = y - 1,\qquad |y - 2| = y - 2.$$

Substituting yields

$$y(y - 1) + 2 = (y - 1) + (y - 2) = 2y - 3.$$ $$y^2 - y + 2 = 2y - 3.$$ $$y^2 - 3y + 5 = 0.$$

The discriminant is $$(-3)^2 - 4(1)(5) = 9 - 20 = -11 < 0,$$ again giving no real root.

Combining all three cases, the only positive $$y$$ that satisfies the equation is

$$y = \frac{\sqrt5 - 1}{2}.$$

Now revert to the original variable $$x$$ using $$y = 3^x$$:

$$3^x = \frac{\sqrt5 - 1}{2}.$$

Taking logarithm base 3 on both sides (or using natural logs divided by $$\ln 3$$) gives the single real solution

$$x = \log_3\!\Bigl(\tfrac{\sqrt5 - 1}{2}\Bigr).$$

Therefore the solution set $$S$$ contains exactly one real number; it is a singleton.

Hence, the correct answer is Option B.